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Question: The equation of second degree \(x^{2} + 2\sqrt{2}xy + 2y^{2} + 4x + 4\sqrt{2}y + 1 = 0\) represents...

The equation of second degree

x2+22xy+2y2+4x+42y+1=0x^{2} + 2\sqrt{2}xy + 2y^{2} + 4x + 4\sqrt{2}y + 1 = 0 represents a pair of straight lines. The distance between them is

A

4

B

4/34/\sqrt{3}

C

2

D

232\sqrt{3}

Answer

2

Explanation

Solution

Distance =2g2−aca(a+b)=24−11(1+2)=2= 2\sqrt{\frac{g^{2} - ac}{a(a + b)}} = 2\sqrt{\frac{4 - 1}{1(1 + 2)}} = 2.